Question 530447

Joe has a collection of nickels and dimes that is worth $7.65. If the number of dimes was doubled and the number of nickels was increased by 35, the value of the coins would be $14.30. How many nickels and dimes does he have?


Let amount of dimes be D, and amount of nickels, N


Then: . 1(D) + .05(N) = 7.65 ------ 10D + 5N = 765 ----- Multiplying by 100 to clear decimals ----- eq (i)


Also, .1(2D) + .05(N + 35) = 14.3 ----- .2D + .05N + 1.75 = 14.3 ------- .2D + .05N = 12.55 ------- 20D + 5N = 1,255 ----- Multiplying by 100 to clear decimals ----- eq (ii)


10D + 5N = 765 ------- eq (i)
20D + 5N = 1,255 ----- eq (ii) 
---------------------------------
- 10D = - 490 ----- Subtracting eq (ii) from (i)


D, or amount of dimes = {{{(- 490)/- 10}}}, or {{{490/10}}}, or {{{highlight_green(49)}}}


10(49) + 5N = 765 ------- Substituting 49 for D in eq (i)
490 + 5N = 765
5N = 765 – 490
5N = 275


N, or amount of nickels = {{{275/5}}}, or {{{highlight_green(55)}}}


I'll leave the check up to you to perform.


Send comments and “thank-yous” to “D” at  MathMadEzy@aol.com